Respuesta :
A- The maximum vertical distance from the water surface to the diver’s
shoulders are 12.061 meters.
B- The time that the diver’s shoulders enter the water is 1.936 seconds.
C- The total distance traveled by the diver’s shoulders from the time
she leaps from the platform until the time her shoulders enter the water is 12.946 meters.
D- The angle, θ between the path of the diver and the water at the instant the diver’s shoulders enter the water is 1.519.
A-
[tex]\frac{dy}{dx} =0[/tex] only when [tex]t=0.36735[/tex]. Let [tex]b=0.36735[/tex]
The maximum vertical distance from the water surface to the diver’s
shoulders is y(b)=11.4 + [tex]\int\limits^b_0 \frac{dy}{dT} dt = 12.061[/tex] meters.
B-
[tex]y(A)=11.4+\int\limits^A_0 \frac{dy}{dt} dt = 11.4+3.6A- 4.9A^{2} =0[/tex]
when A= 1.936 seconds.
C-
[tex]\int\limits^A_0 {\sqrt{(\frac{dx}{dt} )^{2} + (\frac{dy}{dt} )^{2} } } \, dt = 12.946 m[/tex]
At time A, dy/dx = [tex]\frac{dy/dt}{dx/dt} | _{t=A} = -19.21913[/tex]
D- The angle between the path of the diver and the water is [tex]tan^{-1} (19.21913) = 1.518 or 1.519[/tex]
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